Why can't I access a pointer to pointer for a stack array?

Question

Please take a look at the following code. It tries to pass an array as a char** to a function:

#include <stdio.h> #include <stdlib.h>  static void printchar(char **x) {     printf("Test: %c\n", (*x)[0]); }  int main(int argc, char *argv[]) {     char test[256];     char *test2 = malloc(256);      test[0] = 'B';     test2[0] = 'A';      printchar(&test2);            // works     printchar((char **) &test);   // crashes because *x in printchar() has an invalid pointer      free(test2);      return 0; } 

The fact that I can only get it to compile by explicitly casting &test2 to char** already hints that this code is wrong.

Still, I'm wondering what exactly is wrong about it. I can pass a pointer to a pointer to a dynamically allocated array but I can't pass a pointer to a pointer for an array on the stack. Of course, I can easily work-around the problem by first assigning the array to a temporary variable, like so:

char test[256]; char *tmp = test; test[0] = 'B'; printchar(&tmp); 

Still, can someone explain to me why it doesn't work to cast char[256] to char** directly?

Answer 1

test is an array, not a pointer, and &test is a pointer to the array. It is not a pointer to a pointer.

You may have been told that an array is a pointer, but this is incorrect. The name of an array is a name of the entire object—all the elements. It is not a pointer to the first element. In most expressions, an array is automatically converted to a pointer to its first element. That is a convenience that is often useful. But there are three exceptions to this rule:

• The array is the operand of sizeof.
• The array is the operand of &.
• The array is a string literal used to initialize an array.

In &test, the array is the operand of &, so the automatic conversion does not occur. The result of &test is a pointer to an array of 256 char, which has type char (*)[256], not char **.

To get a pointer to a pointer to char from test, you would first need to make a pointer to char. For example:

char *p = test; // Automatic conversion of test to &test[0] occurs. printchar(&p);  // Passes a pointer to a pointer to char. 

Another way to think about this is to realize that test names the entire object—the whole array of 256 char. It does not name a pointer, so, in &test, there is no pointer whose address can be taken, so this cannot produce a char **. In order to create a char **, you must first have a char *.

Answer 2

Because test is not a pointer.

&test gets you a pointer to the array, of type char (*)[256], which is not compatible with char** (because an array is not a pointer). This results in undefined behavior.